Probability of Multiple Failures
Date: 06/20/2002 at 21:56:14 From: Marzineh Farvon Subject: Complete Probability in Calculus If a car has 2 headlights with an average lifespan of 2500 hours, whose probability of failure can be modeled by: f(t) = 0 if t < 0 ; f(t) = 1/u e ^(-t/u) if t >= 0; for the mean value u = 2500, what is the probability that both lamps will fail within 2500 hours, and what is the probability that both will fail in a total of 2500 hours?
Date: 06/21/2002 at 07:27:50 From: Doctor Anthony Subject: Re: Complete Probability in Calculus The probability that either will fail within 2500 hours is 2500 probability = INT [(1/u)e^(-t/u).dt] 0 = (-u/u)e^(-t/u) from 0 to 2500 = -[e^(-1) - e^0] = 1 - e^(-1) = 0.63212 Assuming that multiple failures would be independent events, the probability that BOTH fail within 2500 hours probability = 0.63212^2 = 0.399576 To determine the probability that both will fail within a total of 2500 hours requires a little bookwork. Using 'a' as the parameter the probability distribution function (PDF) of the exponential distribution is f(x) = ae^(-ax) and similarly g(y) = ae^(-ay) Let z = x + y so y = z - x Let S(z) be the cumulative distribution function (CDF) of z = x + y. S(z) = Prob(Z < z) = Prob(X+Y < z) = INT [INT [f(x) g(y) dx dy]] = INT [f(x)dx INT [g(y) dy]] and differentiating S(z) with respect to z we obtain S'(z) = s(z) = INT [f(x) g(z-x) dx] z s(z) = INT [dx ae^(-ax) ae^(-a(z-x))] 0 z = a^2 INT [dx e^(-ax) e^(-az) e^(ax)] 0 z = a^2 INT [dx e^(-az)] (Note: z is constant) 0 = a^2 e^(-az) z The PDF of z is s(z) = a^2 z e^(-az) z >= 0 For this problem we have a = 1/2500 and we require 2500 a^2.INT [z e^(-az) dz] 0 and integrating by parts, = a^2 [z (-e^(-az)/a) + (1/a) INT [e^(-az) dz]] = a [-z e^(-az) - (1/a)e^(-az)] from 0 to 1/a = a[-(1/a)e^(-1) - (1/a)e^(-1) + (1/a)e^(0)] = -e^(-1) - e^-1 + 1 = 1 - 2.e^(-1) = 0.264241 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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